$[0,1)$ is in both $G_{\delta}$ and $F_{\sigma}$

Question

I know $G_{\delta}$ is the complement of $F_{\sigma}$ and it can be proved easily by by using the countable intersection of sets is the complement of the countable union of the complements of the sets.

However, (This might be a very dumb question...) Since I also know $[0,1) = \bigcap^{\infty}_{n=1}(- \frac{1}{n} , 1) = \bigcup^{\infty}_{n=1} [0, 1 - \frac{1}{n}]$ is in both $G_{\delta}$ and $F_{\sigma}$, I wonder how one element can be in a set and its complement.

Thank you in advance!

Answer 1

$G_\delta$ is not the complement of $F_\sigma$. The complement of an $F_\sigma$ set is a $G_\delta$ set.

Answer 2

The complement of a $G_{\delta}$ set is an $F_{\sigma}$ and vice versa, but how does that stop a set from being both an $G_{\delta}$ and an $F_{\sigma}$? In fact every open set (and every closed set) is an $G_{\delta}$ and an $F_{\sigma}$!